已知x=[a+(a^2-1)^(1/2)]^[(2mn)/(m-n)]化简：[x^(1/m)+x^(1/n)]^2-4a^2x^(1/m+1/n)
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解：由x=[a+(a^2-1)^(1/2)]^[(2mn)/(m-n)]得
x^((m-n)/mn)=[a+(a^2-1)^(1/2)]^2
x^(1/n-1/m)=[a+(a^2-1)^(1/2)]^2
x^(1/n)=[a+(a^2-1)^(1/2)]^2*x^(1/m)
则
[x^(1/m)+x^(1/n)]^2-4a^2x^(1/m+1/n)
=[x^(1/m)+[a+(a^2-1)^(1/2)]^2*x^(1/m)]^2-4a^2[a+(a^2-1)^(1/2)]^2*x^(2/m)
=x^(2/m)[[1+[a+(a^2-1)^(1/2)]^2]^2-4a^2[a+(a^2-1)^(1/2)]^2]
=x^(2/m)[1+2[a+(a^2-1)^(1/2)]^2+[a+(a^2-1)^(1/2)]^4-4a^2[a+(a^2-1)^(1/2)]^2]
=x^(2/m)[1+[a+(a^2-1)^(1/2)]^2[2+[a+(a^2-1)^(1/2)]^2-4a^2]]
=x^(2/m)[1+[a+(a^2-1)^(1/2)]^2[2+a^2+2a(a^2-1)^(1/2)+a^2-1-4a^2]]
=x^(2/m)[1+[a+(a^2-1)^(1/2)]^2[1+2a(a^2-1)^(1/2)-2a^2]]
=x^(2/m)[1-[a+(a^2-1)^(1/2)]^2[a-(a^2-1)^(1/2)]^2]
=x^(2/m)[1-[a^2-(a^2+1)]^2]
=x^(2/m)(1-1)^2
=0
x^((m-n)/mn)=[a+(a^2-1)^(1/2)]^2
x^(1/n-1/m)=[a+(a^2-1)^(1/2)]^2
x^(1/n)=[a+(a^2-1)^(1/2)]^2*x^(1/m)
则
[x^(1/m)+x^(1/n)]^2-4a^2x^(1/m+1/n)
=[x^(1/m)+[a+(a^2-1)^(1/2)]^2*x^(1/m)]^2-4a^2[a+(a^2-1)^(1/2)]^2*x^(2/m)
=x^(2/m)[[1+[a+(a^2-1)^(1/2)]^2]^2-4a^2[a+(a^2-1)^(1/2)]^2]
=x^(2/m)[1+2[a+(a^2-1)^(1/2)]^2+[a+(a^2-1)^(1/2)]^4-4a^2[a+(a^2-1)^(1/2)]^2]
=x^(2/m)[1+[a+(a^2-1)^(1/2)]^2[2+[a+(a^2-1)^(1/2)]^2-4a^2]]
=x^(2/m)[1+[a+(a^2-1)^(1/2)]^2[2+a^2+2a(a^2-1)^(1/2)+a^2-1-4a^2]]
=x^(2/m)[1+[a+(a^2-1)^(1/2)]^2[1+2a(a^2-1)^(1/2)-2a^2]]
=x^(2/m)[1-[a+(a^2-1)^(1/2)]^2[a-(a^2-1)^(1/2)]^2]
=x^(2/m)[1-[a^2-(a^2+1)]^2]
=x^(2/m)(1-1)^2
=0
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